avicenna

I think I will search for what Swansont says about "loop" detector.

My problem is that my "aperture area" is very small, say 2x2cm view! I have little space to work with. I know of mini solar cells, but they dont detect IR waves. 

Can you recommend a specific make of mini IR sensors. Say can detect 10,000nm IR. The sensor must not need external voltage source;, just two leads.

Loop antenna is too much for me to build!!!

Say I have created a strong enough micro and ir radiation source from a certain direction. I want a mini photocell/sensor to detect it. 

I know there are many mini photo diodes, etc available and cheap. But I need the sensor to be directional, having only one face to detect the radiation. If I flip the face, there would not be radiations detected. Also it should be easy to use, say just two leads to detect voltage/current.

Also, what electrical instruments do I need to detect a voltage/current. Can a common multimeter do the job.
 

3 minutes ago, exchemist said:

No, spectral lines have finite width for a variety of perfectly good reasons. (Finite line width means there is a range of absorbing or emitting frequencies of course.) These include the Doppler effect, from motion of the emitters relative to absorbers, and uncertainty broadening, due to finite lifetime of the excited state leading to uncertainty in its energy, by Heisenberg's Uncertainty Relations. In gases, this can be a function of pressure, cf. "pressure broadening", since collisions may shorten lifetimes of excited states and also alter their energy, due to transient proximity of second atoms, thereby disturbing the potential experienced by the electrons. And for matter in condensed states, atomic lines tend to get broadened into bands anyway, due to the overlay of vibrational and/or rotational fine structure.

If you read up a bit about spectroscopy, there is quite a bit to it besides simple line emission and absorption. 

Thanks. OK. The world of knowledge is vast as the universe itself. 

It is commonly stated in QM that a bound electron may only absorb a photon if only there is a matching energy gap difference that matches the photon energy. This seems ridiculous.

Say a photon emitted by a hydrogen atom with the typical red emission frequency of the Blamer series. If this photon meets a piece of copper, it is unlikely that a copper atom has an exact matching two energy levels matching that of the Balmer spectrum of hydrogen. The photon would just cruise through copper without being absorbed. Well, if a statement is flawed, we then decide to "patch up" our theory and say there are other means that a photon may be absorbed by matter.

So the scientific method nowadays would be a series of ad hoc "patching ups".      
 

@Carrock, I think your analysis is acceptable. Thanks.

15 hours ago, Carrock said:

There's been a bit of conflation in some replies between A.C. and D.C..  D.C. is conserved but A.C. isn't.

So: V.H.F. A.C. whose effects are more obvious than 50Hz.

You could have 100 meters of 50 ohm coaxial cable with 10db loss per 100 meters. Connecting A.C. power at say 100MHz 100V to the (resistive) cable gives I =V/R = 2 amps input (200W). Connect a 50 ohm resistive load to the far end and you'll get ~ 31.6V at 0.632A i.e. 20W output. If you use a 200 meter cable you'll get 10V at 0.2A i.e. 2W. The input is still 200W into 50 ohms.

The current drops exponentially along the cable. The main power losses are I^R losses in the conductor, dielectric (insulator) heating and radiation from the cable. There is no A.C. current conservation; some of it charges and discharges the dielectric and current is also involved in creating magnetic and electromagnetic fields.

One way of dealing with reactance is to consider the effect of a load impedance mismatch.

e.g. terminate the cable with 25 instead of 50 ohms. This will cause a power reflection back into the cable to compensate for trying to connect a 50 ohm cable to a 25 ohm load. If 20W output then 20 *(50 -25)/(50+25)W i.e. 6.7W is reflected back into the cable and after attenuation 0.67W reaches the source.

You'll get standing waves on the cable; every half wavelength (About 1.5m) you'll get maximum voltage and minimum current; between these nodes you get a minimum voltage, maximum current node. A.C. current can be created and destroyed without breaking conservation laws. Some energy is stored in various fields and doesn't reach the load; sometimes it's called imaginary power(it can be treated as 90 deg or sqrt(-1) out of phase with 'real' power) or reactive power(capacitors and inductors have reactance).

This is sort of real; there are meters which measure forward and reflected power in coax cables...

I didn't want to oversimplify too much; this post ended up much longer than I intended.

I believe this is the real deal, but only for those with some foundations in AC circuit theory. I will see how things relate to what I am investigating.  

@Carrock. 

I am trying to reduce AC circuit to DC circuit "instantaneously". 

I don't want coaxial cables which complicates things. Say I have a simple ac generator that generates fairly good sinusoidal voltage source at 50 Hz (If possible at all?). I connect a long resistive wire to the terminals in a huge circular loop. When the wire is at thermal equilibrium with the environment, we know that all power will be dissipated as IR radiation loss, purely resistive loss - assume ideally.

So we could always apply ohm's law of I=V/R where R is the resistance of the wire, V the instantaneous voltage. It seems that there will be the usual charge conservation along the wire as if the current is a dc current. The current should be a constant at that moment of consideration.

My setup would eliminate capacitance, inductance etc. Instantaneously, we only have the magnetic fields around the wire which we assume "steady". How is such an analysis.
 

2 minutes ago, swansont said:

The latter part of my explanation applies to this. The current is uniform/\. There’s no way for it to vary.

Thanks. This is what I want to confirm.

For the AC case, I really cannot say as the real world and the ideal world may be different. 

3 minutes ago, swansont said:

If the voltage is constant (for a real wire) there would be no current. If there is a voltage drop, and thus a current, the current will be uniform even if the voltage drop is not (e.g. if there’s a resistor, or a series of different-valued resistors); there’s no way to vary it. Charge is conserved, so current flowing in to a point equals the current flowing out.

I am saying the potential difference between the ends of the wire, i.e. a long wire connected to a dc battery.

36 minutes ago, sethoflagos said:

I think the correct answer to this rather odd question is yes, though if the wire has any resistance then that uniform current is zero. In the absence of reactive elements (capacitors or inductors) a constant current implies a constant voltage gradient.

I don't  understand. I = V/R, so how can there be zero current if we connect a AA 1.5 V battery to a long copper wire. My question is whether I is the same at all points of the wire.  

If the voltage across a long wire is constant, is the current uniform throughout the wire length.  

In a 50 Hz ac voltage across a long wire, at a certain moment how does current vary along x, the wire length. 
 

Maybe no easy answer because we know too little yet about light.

The wiki says it is the bound electrons that absorbs radiation; then how the electrons energy get transferred  to the nucleus kinetic energy. I think temperature not dependent on the KE of electrons, but only in the KE of the nucleus or center of mass.  

I'll like to know the actual physical mechanism how matter absorbs EM radiation. Take the specific example how copper metal absorbs IR radiation and the copper having its temperature raised.
 

In a capacitor discharge through a plain resistor, the capacitor power supplied at any instant is VI; the power dissipated in the resistor is I²R. So VI = I²R.

Consider a railgun operated with a capacitor bank. At any instant of capacitor discharge, the power supplied is VI. The total power supplied for ohmic loss is sum I²R for two rails plus the resistance of the armature.

Question: Since VI = total I²R, how can the power equation include the kinetic energy supplied to the armature?